+0

# 0.4^((6-5x)/(2+5x))<(25/4)

0
458
5

0.4^((6-5x)/(2+5x))<(25/4)

How could I solve this :) ? thanx

Guest May 17, 2014

#4
+91510
+8

hi Blaster0

***How did you get -2 from log6,25/log0,4***

To be perfectly honest I just put the whole thing straight into a calculator and it spat out -2

However, since I am on a roll here I will explain why it is so!

$$\frac{log6.25}{log0.4}=log_{0.4}6.25$$     I did this by applying the change of base law the opposite way around to usual.

Let

$$x=log_{0.4}6.25$$

Rearranging this we get

$$6.25=0.4^x\\ 2.5^2=0.4^x\\ \left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^x\\ \left(\frac{2}{5}\right)^{-2}=\left(\frac{2}{5}\right)^x\\ x\:=\:-2$$

Hey Blaster0,

When you are happy with your answer can we have a thank you and some thumbs up please.

Speaking for myself I put a lot of effort into these answers!

(Don't mind me I am just trying to train people to be polite.  I figure that the more people who do it then the more normal it becomes and hopefully the 'idea' will catch on.  )

Melody     And no chris you can not have your glasses back - I like them!

Melody  May 18, 2014
Sort:

#1
+91510
+8

this should be the last one for the night - It is 1:30 am

$$0.4^{((6-5x)/(2+5x))}<6.25\\\\ log0.4^{((6-5x)/(2+5x))} \frac{6-5x}{2+5x}log0.4 \frac{6-5x}{2+5x}>\frac{log6.25}{log0.4}\\\\ \frac{6-5x}{2+5x}>\frac{log6.25}{log0.4}\\\\ \frac{6-5x}{2+5x}>-2\\\\ (2+5x)^2\frac{6-5x}{2+5x}>-2(2+5x)^2\\\\ (2+5x)(6-5x)>-2(25x^2+20x+4)\\\\ 12-10x+30x-25x^2>-50x^2-40x-8\\\\$$

$$25x^2+60x+20>0\\\\ 5x^2+12x+4>0\\\\ 5x^2+10x+2x+4>0\\\\ 5x(x+2)+2(x+2)>0\\\\ (5x+2)(x+2)>0\\\\ \mbox{concave up parabola roots are -2 and -2/5}\\\\ x<-2 \mbox{ OR }x>\frac{-2}{5}$$

Okay I decided to check this by plotting the original question.  And to my delight it appears to be correct!

Melody  May 17, 2014
#2
+38
+5

How did you get -2 from log6,25/log0,4

thx

blaster0  May 17, 2014
#3
+81123
+8

$${log}_{10}\left({\mathtt{6.25}}\right) = {\mathtt{0.795\: \!880\: \!017\: \!344\: \!075\: \!2}}$$

$${log}_{10}\left({\mathtt{0.4}}\right) = -{\mathtt{0.397\: \!940\: \!008\: \!672\: \!037\: \!6}}$$

If you divide the first by the second, the result is -2.

CPhill  May 17, 2014
#4
+91510
+8

hi Blaster0

***How did you get -2 from log6,25/log0,4***

To be perfectly honest I just put the whole thing straight into a calculator and it spat out -2

However, since I am on a roll here I will explain why it is so!

$$\frac{log6.25}{log0.4}=log_{0.4}6.25$$     I did this by applying the change of base law the opposite way around to usual.

Let

$$x=log_{0.4}6.25$$

Rearranging this we get

$$6.25=0.4^x\\ 2.5^2=0.4^x\\ \left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^x\\ \left(\frac{2}{5}\right)^{-2}=\left(\frac{2}{5}\right)^x\\ x\:=\:-2$$

Hey Blaster0,

When you are happy with your answer can we have a thank you and some thumbs up please.

Speaking for myself I put a lot of effort into these answers!

(Don't mind me I am just trying to train people to be polite.  I figure that the more people who do it then the more normal it becomes and hopefully the 'idea' will catch on.  )

Melody     And no chris you can not have your glasses back - I like them!

Melody  May 18, 2014
#5
+38
+5

I got it

Melody and CPhill thank you very much and sorry for that thumbs

blaster0  May 18, 2014

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